Download PHY140Y 2 Measuring G and Gravitational Potential Energy

PHY140Y
Measuring G and Gravitational Potential Energy
2
2.1
Overview
• Measurement of G
• Gravitational Potential Energy
• The Truth About Escape Velocities
2.2
Measuring G
When Newton put forward his law of universal gravitation aroung 1670, he was not able to give a
precise value for the constant of universal gravitation G. This had to wait over a hundred years till
1798, when Lord Henry Cavendish made the first accurate determination of G.
Why is it so difficult to measure G? Think again about the nature of gravity. We are aware
of it because of our own experience of its force on the surface of the Earth and we can accurately
measure the acceleration of gravity. But from the law of universal gravitation, what we know is
g =
GME
2 .
RE
(1)
We see that if we know the radius of the Earth, RE , we can only infer GME , namely the G times
the mass of the Earth. Similarily, knowing the moon’s orbit, it’s radius and an estimate for it’s
mass, we only infer GME . A direct measurement of the force of gravity, using two masses of know
size is required. And because gravity is so weak, designing a sufficiently sensitive instrument is a
challenge.
Cavendish addressed this by developing what we call the “Cavendish Balance,” shown in Fig. 1.
He fastened two 5-cm lead balls together with a rod, and then fastened the rod to a thin fibre that
he used to suspend them. He then brought two large spheres in proximity to each of the two lead
spheres on the “dumbell.” The force of gravity caused the masses to attract one another, twisting
the fibre. By knowing the torsion coefficient on the fibre, he was able to estimate a value for the
force. This allowed him to directly calculate G.
The most precise estimate of G is
G = (6.67259 ± 0.00085) × 10−11 N m2 /kg2 .
(2)
The precision to which this is measured, about 130 parts in a million, is not outstanding. It turns
out to be a difficult constant to measure because it is so small. Note the units of G!
1
Quartz
Fibre
Pb Mass
-Fg
Laser
Source
Mirror
Dumbell
Fg
Beam
Spot
Recorder
Pb Mass
Figure 1: A schematic of the Cavendish Balance, as seen in our first and second-year laboratory.
2.3
Gravitatational Potential Energy
Once we have defined a force of gravity, the next natural step from a dynamics point of view is
to introduce the concept of energy into the picture. We have already seen that in the context of
a constant acceleration due to gravity, g, (which is only true over distances small compared to the
radius of the object doing the attracting), we can define a gravitational potential energy
U ≡ mgh,
(3)
where m is the mass of the object and h is the distance above some reference point where we define
U ≡ 0.
To generalize this, we start with a massive object with mass M , and find a way of calculating
the work done to move an object in its gravitational field from one place to another. It is clear
that if we move an object and keep it at the same proximity or radius from the large mass, we do
not gravitational work as whatever motion has to take place happens at right angles to the force
of gravity. Thus, we would only do work with we move the object in radius. So, suppose we have
to move the object from point r1 to point r2 in the gravitational field. We then find that the work
we have to do is given by
r2
W
=
r1
= −
F · dr
r2
GM m
r1
2
r2
(4)
dr
(5)
=
GM m
r
r2
(6)
r1
1
1
−
.
= GM m
r2 r1
(7)
We see that the work we have to do only depends on the initial and final radii of the mass m, which
tells us that it doesn’t matter which we path we chose to move the mass.
So the work we do to move the object goes into increasing the object’s gravitational potential
energy, U . We therefore can relate W to the negative of the change in the object’s potential energy:
1
1
−
U (r2 ) − U (r1 ) = −GM m
r2 r1
GM m
U (r2 ) = −
+ constant,
r2
(8)
(9)
where the constant in the right hand side is determined by our choice of r1 . This choice is in fact
totally arbitrary, since in principle, it is only the change in potential energy that has any physical
significance, because it is that change that we relate to the work performed on the object (and
therefore the forces acting on it).1 It is simplest and most intuitive to chose r1 = ∞ as that means
the constant is zero and we can ignore it.2
This is now the generalization of our original gravitational potential energy. We now see that
g=
GME
,
RE
(10)
where ME = 5.97 × 1024 kg is the mass of the Earth and RE = 6.37 × 106 m is the radius of the
Earth (check it out!).
2.4
Escape Velocity
Let’s suppose I am sitting out in space at an initial radius ri far away from the Earth (let’s assume
ri >> RE ). If I start from at rest, the I will begin to accelerate toward the Earth under the force
of gravity. If we assume that air resistance is neglible, then I will end up impacting the surface of
the Earth with a considerable velocity. Let’s work out the details.
My total energy when we start this is
Ei = K + U
GME mp
,
= 0−
ri
(11)
(12)
where mp is my mass (which I claim shouldn’t matter). When I impact the Earth, my total energy
will still be the same (remember, we are ignoring important effects such as air resistance). Thus,
1
Can you think of a counter-example to this? See if there is any simple system where the absolute value of the
potential energy makes any difference to what you would observe.
2
This is the only general choice that makes any sense to me, since r1 = 0 is not a possibility and any other value
of r1 would be totally arbitrary.
3
my velocity on impact, vf , will be given by conservation of total energy,
Ei = E f
1
⇒ mp vf2
2
⇒ vf
1
GMe mp
mp vf2 −
2
RE
1
1
= GMe mp
−
RE
ri
=
=
2GM
RE
(13)
(14)
1−
RE
.
ri
(15)
Since the distance I start from, ri , is so much larger than RE , I can ignore the term in the parentheses
and arrive at an expression for my impact velocity
vf
=
2GM
= 1.12 × 104 m/s = 11.2 km/s.
RE
(16)
This velocity is usually referred to as the “escape velocity,” vesc , because it would be the
velocity at which I would have to leave the surface of the Earth in order to “escape” from the
Earth’s gravitational field. In this sense, it is a curious number as it is not really relevant when we
want to lauch a rocket.
You can see this by asking how long, under a reasonable acceleration like a = 10 m/s2 , would
it take for something to reach this escape velocity. It would take a time
∆t =
vesc
∼ 1, 120 s,
a
(17)
and during that time it would travel a distance
∆d =
1
a (∆t)2 ∼ 6.3 × 106 m.
2
(18)
No rocket ever behaves in this way! What we normally do is first accelerate a rocket with a
somewhat larger acceleration (3-4 g’s is about typical) for a period of several minutes to get it into
a low Earth orbit of order 500 km in one stage and then take it out of Earth orbit with a second
stage that allows it to achieve the necessary velocity to take it out of Earth’s gravitational field. It
is at this stage where much higher velocity’s are attained, in part because one can throw away the
first stage of the rocket that got the vehicle out of the Earth’s atmosphere.
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